Schemes for Video on demand

Yuan-Shiang Yeh

Outline

Introduction Previous Works Study

Buffer Requirement Channel Adjustment Bandwidth reduction in multi-layer

videos

introduction

True VoD Batch Patch

Near VoD Fast Data Broadcasting Harmonic Boradcasting

Comparison

True VoD Near VoD

Delay Time no yes( 依方法不同 )

Receive Type Peer to Peer Broadcast

Client number dependent independent

Suitable Video

any hot

Objective Bandwidth

Ex: FB -> Pagoda -> New Pagoda Buffer Storage

Ex: PB -> PPB Scalable (wide range)

Ex: FB -> UD VCR Functionality

Ex: Staggered Broadcast -> SAM Channel Adjustment

Ex: FB -> Seamless Channel Transition I/O BANDWIDTH

Ex: FB -> SB

Harmonic Broadcasting Scheme

Parameters: Movie length --- D (e.g., 120 minutes) Consumption rate of the movie --- b (e.g., 10Mb

ps) Size of the movie --- S = D*b The movie is equally divide into N segments, an

d Si is the ith segment of the movie. Viewer waiting time --- d

d = D / ND

S1 S2 S3 S4

d

bandwidth

Harmonic Broadcasting Scheme

Parameters The ith segment of the movie Si is equally

divided into i sub-segment(s) {Si, 1, Si, 2 --- Si, i}

Let the i sub-segment(s) of Si be put on a logical channel Ci, the bandwidth of Ci is b/i.

Harmonic Broadcasting Scheme

The total bandwidth(B) allocated for the movie is as follows:

Where HN is called the harmonic number of N

B = b + b/2 + b/3 + b/4 = 2.083b

HN = 1 + 1/2 + 1/3 + 1/4 = 2.083

S3, 1S2, 1

S1 S3, 2

S3, 3

S2, 2

S4, 1

S4, 2

S4, 3

S4, 4

D

B

S1 S1 S1 S1

S2, 1 S2, 2 S2, 1 S2, 2

S3, 1 S3, 2 S3, 3 S3, 1

S4, 1 S4, 2 S4, 3 S4, 4

C1 B

C2 2B

C3 3B

C4 4B

d

Time

Bandwidth

S1

S2, 1

S3, 1

S4, 1

S2, 2

S3, 2

S4, 2

S2, 1

S3, 3

S4, 3

S3, 1 S3, 2

S4, 4S4, 1 S4, 2 S4, 3

Fast broadcasting scheme(FB)

In FB scheme, we divide a movie into 2k - 1 segments, k channels is needed.

Waiting time: d = D / 7 ( D: the video length )

S1

S2

S4

S1

S3

S5

S1

S2

S6

S1

S3

S7

S1

S2

S4

S1

S3

S5

S1

S2

S6

S1

S3

S7

bChannel 1

Channel 2

Channel 3

S1

D

d

S2 S7

C1

C2

C3

· · ·S1 S1 S1 S1 S1 S1 S1 S1

S2 S3 S2 S3 S2 S3 S2 S3

S4 S5 S6 S7 S4 S5 S6 S7

· · ·

· · ·

S2 S2 S2 S2

S1

S2

S4

S3

S5

S2

S6

S3

S7S4

Pagoda Broadcasting Scheme

C1

C2

C3

C2k

C2k+1

…

Segment Sz needs to be transmitted at minimum frequency 1/(zd)

Pagoda Broadcasting Scheme

Segments Stream Broadcasting Frequency

S10 to S14 4 1/10d

S15 to S19 5 1/15d

S20 to S29 4 1/20d

S30 to S49 5 1/30d

4z = 40

Example : kou’s researchSegment number: 7×2i, i = 0, 1, … 5;Number of channels: i + 4;Client buffer size: 2i segments;

S1C0

t0

d

C0 S1

t0 + d

S2C1

S1C0 S1

t0 + 2d

S2C1

S1

S3

Playing segment

Buffered segment

V0 S1

V0 S2V1 S1 S2

V0 S3V1 S2 S3V2 S1 S3

S1

No

No

No

S1C0 S1

t0 + 3d

S2C1

S1

S3

S1

S2

S4C2

S1C0 S1

t0 + 4d

S2C1

S1

S3

S1

S2

S4C2

S1

S5

S6

C3 S4

V0 S4V1 S3 S4V2 S2 S3V3 S1 S2

V0 S5V1 S4 S5V2 S3 S4V3 S2 S4V4 S1 S4

S6

No

S1C0 S1

t0 + 7d

S2C1

S1

S3

S1

S4

S2C2

S1

S5

S6

S4

S1

S2

S3

S7

S1

S5

S6

S4

S2

S3

S7

S1C0 S1

t0 + 6d

S2C1

S1

S3

S1

S4

S2C2

S1

S5

S6

C3 S4

S1

S2

S3

S7

S1

S5

S6

S4

Buffer Requirement - FB

1

24

9

13510

12611

13712

12413

13514

126

8

137

124

135

126

137

124

135

126

137

C1

C2

C3

C4 15 9 10 11 12 13 148 15

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

248

48

3

95

48

95

106

8

95

106

117

89

106

117

12

8910

117

1213

891011121314

91011121314

1011121314

11121314

121314

1314

14

Current segment

Buffer

Buffer Requirement - FB

1

9

13510

12611

13712

12413

13514

126

13715

248

7 0 1 2 3 4 5 6

28

Buffer Requirement = 7/15

Our permutation

1249

13611

12513

13715

1248

13610

12512

13714

1249

13611

12513

13715

1248

13610

12512

13714

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

2 3 48

6810

681012

781012

891012

910

1211

10

1211

13 15

1211

13 15

1213 15

13 15 15

1249

C1

C2

C3

C4

Current segment

Buffer

Buffer Requirement - ours

1

10

13612

12514

137

9

12411

13613

125

13715

248

4 3 3 3 3 4 4 4

28

Theorem 1:Assuming that a video is equally divided into segments and allocated to n channels as Fast Data Broadcast did, our broadcasting schedule requires the minimum storage space. #

)12( n

Result

channels FB kou oursBuffer Total

segment

Buffer Total segmen

t

Buffer Total segmen

t

4 2 7 1 7 1 7

5 4 15 2 14 2 15

6 8 31 4 28 4 31

7 16 63 8 56 8 63

8 32 128 16 112 16 128

Bandwidth - Algorithm 1 1. Divide channels into slots. Channel i

would broadcast segment i. 2. For current segments k, find the slo

t with minimum wasted bandwidth. 3. Suppose the slot is divided into x su

bslots ,then allocate the segment k to (x+k-1) to this slot.

4. Repeat 2~3 until no free slots.

1. Divide into slots

C1

C2

C3

S1

S3

S2

free free

Broadcasting frequency = 1

free

2. Find suitable slot

C1

C2

C3

S1

S3

S2

free free

free

Current segment = S4

free free

1/2 * 1/2 - 1/4 = min

3. Allocate segments

C1

C2

C3

S1

S3

S2

free free

free

Current segment = S4

free free

1/2 * 1/2 - 1/4 = min

Num of sub-slot = 2

S4 S5

4.Repeat step 2~3

C1

C2

C3

S1

S3

S2

S6

S4 S5

S7 S8 S9

Bandwidth - Algorithm 2

1. Divide free channel or slots with minimum wasted bandwidth for segment i.

2. Allocate segment i into current slots.

3. Repeat 1~2 until no enough free bandwidth.

Example for Algorithm 2

freeC1

C2

C3 free

free

S1

Broadcasting frequency = 1

Example for Algorithm 2

C1

C2

C3

S1

S3

S2 S4 S5

free freeS6 free

Resultchannels

FB PB Algo. 1

NPB Algo. 2

HB

3 7 9 9 9 9 10

4 15 19 24 26 25 30

5 31 49 63 66 73 82

6 63 99 158 172 201 226

7 127 249 412 442 565 615

8 255 499 1056 1522 1673

1

2

4

1

3

5

1

2

6

1

3

7

1

2

4

1

3

5

1

2

6

1

3

7

1

2

4

1

3

5

1

2

6

1

3

7

1

2

4

1

3

5

1

2

6

1

3

7

Base Layer

Enhancement Layer 1

Enhancement Layer 2

1 2

4

3

5 6 7

1 2 43 5 6 7

1 2

4

3

5 6 7

1 2 43 5 6 7 1 2

4 5 6 7 4 5 6 7

1 2 3 1 2 3 1 2

2 channels

1 channels

dummy

popular

Delay in enhancement layer

Delay Slots = k

Frequency (Si) = 1/(i+k)

S1

1

1/2

S2

1/2

1/3

S7

1/7

1/8

Delay slots = 1

Example

C1

C2

C3

Current segment = S1

S1

S2 free free

free freeS3

Current segment = S2Current segment = S3

Result – delay slots

Delay slots

0 1 2 3 4 5 6

Algo.1 158 411 702 1022 1322 1722 2022

Algo.2 201 564 928 1282 1664 1987 2389

Channels = 6 Number of segment in FB scheme = 63